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user100 [1]
4 years ago
8

Help asap!! interpret the meaning of the expression

Mathematics
2 answers:
Triss [41]4 years ago
4 0

Answer:

D

Step-by-step explanation:

This is exponential growth, and to my understanding, the format goes:

<em>initial amount (percent growth/ decay)^time</em>

percent growth = (<em>decimal percent + 1)</em>

<em>percent decay = (1 - decimal percent) </em>

Your equation:

1500(1.02)^t

Using the above format, 1500 appears to be the initial amount, which increases by 2% per annum.

i think

Lilit [14]4 years ago
3 0

Answer D

Step-by-step explanation:

FOR PLATO

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Step-by-step explanation:

6x + c = k ( isolate the term in x by subtracting c from both sides )

6x = k - c ( isolate x by dividing both sides by 6 )

x = \frac{k-c}{6}

5 0
1 year ago
21×9=9×21 what property is used
kvv77 [185]
Hello There =)

Commutative property is used. The commutative property <span>states that two numbers can be multiplied in either order.</span>


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3 years ago
If b is the midpoint of AC, which of the following choices is the value of x?
katrin [286]
If B is the midpoint of AC, that means that the distances from A to B and B to C are the same. The distance from A to B is 8x + 4, and the distance from B to C is 10x - 6, so we can set these two equivalent:

8x + 4 = 10x - 6

Solving for x:

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10 = 2x (add 6 to both sides)
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4 years ago
Which correctly describes a cross section of the square pyramid? Check all that apply.
MariettaO [177]

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acd

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6 0
3 years ago
Read 2 more answers
According to a survey, high school girls average 100 text messages daily (The Boston Globe, April 21, 2010). Assume the populati
Ghella [55]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: number of daily text messages a high school girl sends.

This variable has a population standard deviation of 20 text messages.

A sample of 50 high school girls is taken.

The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;δ²/n)

This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:

Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)

a.

P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836

b.

P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)

P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328

I hope you have a SUPER day!

3 0
3 years ago
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